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Sia ? 6 years, 4 months ago
Here, base, b = 24 cm, and let each equal side be a cm. Then,
area = {tex}\frac { 1 } { 4 } b \sqrt { 4 a ^ { 2 } - b ^ { 2 } } \text { sq units } = \frac { 1 } { 4 } \times 24 \times \sqrt { 4 a ^ { 2 } - 576 } \mathrm { cm } ^ { 2 }{/tex}
={tex}12 \times \sqrt { a ^ { 2 } - 144 } \mathrm { cm } ^ { 2 }{/tex}
But, area = 192 cm2 [given].
{tex}\therefore \quad 12 \times \sqrt { a ^ { 2 } - 144 } = 192 \Rightarrow \sqrt { a ^ { 2 } - 144 } = 16{/tex}
{tex}\Rightarrow a ^ { 2 } - 144 = 256 \Rightarrow a ^ { 2 } = 400 \Rightarrow a = 20{/tex}
{tex}\therefore\quad {/tex}perimeter of the triangle = (2a + b) cm
= (2 {tex}\times{/tex} 20 + 24) cm = 64 cm.
Manoj Kumar 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
Linear pair axiom of theorems are if a ray stands on a line , then the sum of two adjacent angles so formed is 180 degree.
Krishnaansh Viz 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given : ABC is a triangle in which altitude BE and CF to side AC and AB are equal.
To Prove : {tex}\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }{/tex}
- AB = AC i.e. {tex}\triangle{/tex}ABC is an isosceles triangle.
Proof : BE = CF ...... [Given]
∠BAE = ∠CAF ...... [Common]
∠AFB = ∠AFC ....... [Each 90o]
{tex}\therefore{/tex} {tex}\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }{/tex} ........ [By AAS property] - {tex}\triangle A B E \cong \triangle A C F{/tex} ....... [As proved]
{tex}\therefore{/tex} AB = AC . . .[c.p.c.t.]
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC is an isosceles triangle.
Posted by Karan Chauhan 6 years, 5 months ago
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Pratham Jaim 6 years, 5 months ago
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Sia ? 6 years, 4 months ago
Given: AD and BC are equal perpendiculars to a line segment AB.
To Prove: CD bisects AB
Proof : In OAD and OBC
{tex}\angle{/tex}AOD = {tex}\angle{/tex}BOC ...[Vertically opposite angles]
{tex}\angle{/tex}OAD = {tex}\angle{/tex}OBC ...[each 90°]
AD = BC ...[Given]
{tex}\therefore{/tex} DOAD {tex}\cong{/tex} DOBC ...[By AAS property]
{tex}\therefore{/tex} OA = OB ...[c.p.c.t.]
{tex}\therefore{/tex} CD bisects AB
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